rbind for matrices - rep argument

kronecker(rep(1, 6), data.matrix(Xdf))

On 07-Jan-09 15:22:57, Niccol? Bassani wrote:

Dear R users,I'm facing a trivial problem, but I really can't solve it.
I've tried a dozen of codes, but I can't get the result I want.
The question is: I have a dataframe like this one

[,1] [,2] [,3] [,4] [,5]
[1,]    1    2    3    4    5
[2,]    2    5    5    4    9
[3,]    1    6    8    1    2
[4,]    8    6    4    1    5

made up of decimal numbers, of course.
I want to append this dataframe to itself a number x of times, i.e. 3.
That is I want a dataframe like this

[,1] [,2] [,3] [,4] [,5]
[1,]    1    2    3    4    5
[2,]    2    5    5    4    9
[3,]    1    6    8    1    2
[4,]    8    6    4    1    5
[5,]    1    2    3    4    5
[6,]    2    5    5    4    9
[7,]    1    6    8    1    2
[8,]    8    6    4    1    5
[9,]    1    2    3    4    5
[10,]    2    5    5    4    9
[11,]    1    6    8    1    2
[12,]    8    6    4    1    5

I'm searching for an "authomatic" way to do this (I've already used the
rbind re-writing x times the name of the frame...), as it must enter a
function where one argument is exactly the number x of times to repeat
this frame.

Any ideas??
Thanks in advance!
Niccol?

I don't know whether there is anywhere a ready-made function which
will implement a "rep" paramater for an rbind, but the following ad-hoc
function will do it for you efficiently (i.e. with the minimum number
of applications of the rbind() function).

To produce a result which consists of k replicates of x, row-bound:


 Krbind <- function(x,k){
   y <- x
   if(k==1) return(x)
   p <- floor(log2(k))
   for(i in (1:p)){
     z <- rbind(y,y)
     y <- z
   }
   k <- (k - 2^p)
   if(k==0) return(y) else return(rbind(y,Krbind(x,k)))
 }

## Example:

 Xdf <- data.frame(X1=c(1.1,1.2),X2=c(2.1,2.2),
                   X3=c(3.1,3.2),X4=c(4.1,4.2))

 Krbind(Xdf,6)
#     X1  X2  X3  X4
# 1  1.1 2.1 3.1 4.1
# 2  1.2 2.2 3.2 4.2
# 3  1.1 2.1 3.1 4.1
# 4  1.2 2.2 3.2 4.2
# 5  1.1 2.1 3.1 4.1
# 6  1.2 2.2 3.2 4.2
# 7  1.1 2.1 3.1 4.1
# 8  1.2 2.2 3.2 4.2
# 9  1.1 2.1 3.1 4.1
# 10 1.2 2.2 3.2 4.2
# 11 1.1 2.1 3.1 4.1
# 12 1.2 2.2 3.2 4.2

Of course, if you're not worried by efficiency, then the simple loop

 y <- x
 for(i in (1:(k-1))){y <- rbind(y,x)}

will do it!

Hoping this helps,
Ted.

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E-Mail: (Ted Harding) <Ted.Harding at manchester.ac.uk>
Fax-to-email: +44 (0)870 094 0861
Date: 07-Jan-09                                       Time: 18:08:14
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